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CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 10 - MCQExams.com
CBSE
Class 10 Physics
Light Reflection And Refraction
Quiz 10
Which of the following materials has maximum optical density?
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Glass
0%
Water
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Pearl
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Diamond
Explanation
Diamond
Diamond has the highest refractive index of $$2.42$$. The larger the refractive index of a medium, the greater is its optical density.
The magnification of plane mirror is always ...............
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More than 1
0%
1
0%
Less than 1
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Zero
Explanation
1
Since the size of the image formed by a plane mirror is equal to the size of the object, its magnification is always 1.
Which of the following formula is wrong in accordance with Snell's law ?
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$$ \dfrac{n_1}{\sin i} = \dfrac{n_2}{\sin r} $$
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$$ \dfrac{n_{2}}{n_1} = \dfrac{\sin i}{\sin r} $$
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$$n_{1} \sin i = n_{2} \sin r$$
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$$\dfrac{1}{n_2} \sin i = \dfrac{1}{n_1} \sin r $$
Explanation
Snell's Law of refraction states that the product of sine of angle made by the ray in a medium and the refractive index of the medium is constant.
i.e., $$n_{1} \times \sin i = $$ Constant
$$\therefore n_{1} \sin i = n_{2} \sin r $$
$$\Rightarrow \boxed{\dfrac{n_1}{n_2} = \dfrac{\sin r}{\sin i}} $$
Correct Option is A.
A concave mirror forms the real image of an object which is magnified 4 times. The object is moved 3 cm away, the magnification of the image is 3 times. What is the focal length of the mirror?
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3 cm
0%
4 cm
0%
12 cm
0%
36 cm
Explanation
For mirror $$u=\frac {f(m-1)}{m}$$
In first case, $$u=\frac {f(-4-1)}{-4}$$
In the second case, $$u+3=\frac {f(-3-1)}{-3}$$
On solving, we get $$f=36 cm$$
A convex mirror has a focal length of 20 cm. A real object is placed at a distance of 20 cm from the pole of the mirror and in front of the mirror. The mirror produces an image at :
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0%
Infinity
0%
20 cm
0%
40 cm
0%
10 cm
A rod of length 10 cm lies along the principal axis of concave mirror of focal length 10 cm in such a way that
its end closer to the pole is 20 cm away from the mirror. The length of the image is then -
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0%
15 cm
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2.5 cm
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5 cm
0%
10 cm
Explanation
Given for the concave mirror :
Focal Length, $$f=-10cm$$
Length of the rod, $$L_{AB}=10cm$$
Closer end distance of rod, $$u_A=-20cm$$
Far end distance of rod, $$u_B=u_A+L_{AB}$$
$$u_B=-20-10=-30cm$$
By using mirror formula for end $$A$$:
$$\dfrac{1} {V_{A}} + \dfrac{1} {-20} = \dfrac{1} {-10}$$
$$V_{A} = -20 cm$$ [Image position of the end A]
$$\dfrac{1} {V_{B}} + \dfrac{1} {-30} = \dfrac{1} {-10}$$
$$\dfrac{1} {V_{B}} = \dfrac{1} {15}$$
$$v_{B} = -15 cm$$
Length of the image can be given as:
$$L_{A'B'} = \left |V_{A} \right | - \left |V_{B} \right | = 20 - 15 = 5 cm$$
Refractive indices of water and glass are 4/3 and 3/2 respectively. A ray of light travelling in water is incident on the water-glass interface at $$30^0$$. Calculate the sine of angle of refraction.
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0.460
0%
0.585
0%
0.444
0%
0.623
Explanation
Angle of incidence $$i = 30^o$$
Refractive index of water, $$n_w = \dfrac{4}{3}$$
Refractive index of water, $$n_g = \dfrac{3}{2}$$
Let the angle of refraction be $$r$$.
Using Snell's law of refraction :
$$n_w \sin i = n_g \sin r$$
$$\therefore$$ $$\dfrac{4}{3}\times \sin 30^o = \dfrac{3}{2} \sin r$$
Or
$$\dfrac{4}{3}\times 0.5 = \dfrac{3}{2} \sin r$$ $$(\because \sin 30^o = 0.5)$$
$$\implies \sin r = 0.444$$
You are given water, mustard oil, glycerine, and kerosene. In which of these media a ray of light is incident obliquely at same angle would bend the most ?
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Kerosene
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Water
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Mustard oil
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Glycerine
Explanation
Refractive index for water, mustard oil, glycerine and kerosene are:
$$\mu_w = 1.33$$
$$\mu_m = 1.47$$
$$\mu_g = 1.473$$
$$\mu_k = 1.44 $$
Since refractive index of glycerine is highest, ray bends the most in case of glycerine.
A pin that is 2 cm long is placed at a distance of 16 cm from a convex lens. Assuming it to be perpendicular to the principal axis, find the position of the image if the focal length of the lens is 12 cm.
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48 cm
0%
24 cm
0%
12 cm
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None of these
Explanation
Here $$u=-16 cm$$ and $$f=+12 cm$$
We have $$\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {1}{f}$$
or, $$\dfrac {1}{v}=\dfrac {1}{u}+\dfrac {1}{f}$$ $$=\dfrac {1}{-16 }+\dfrac {1}{12 }=\dfrac {1}{48}$$
or, $$v=+48\ cm$$
The image is formed 48 cm from the lens on the side of the transmitted rays. The image is, therefore, real.
What is the focal length of the convex lens ?
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20 cm
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40 cm
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19 cm
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12 cm
The magnitude of focal length of a concave mirror is $$f$$. An object is placed at a distance $$x$$ from the focus and forms a real image. The magnification is:
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$$-\dfrac{f}{x}$$
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$$-\dfrac{x}{f}$$
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$$-\dfrac{f}{f-x}$$
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$$\dfrac{x}{x-f}$$
Explanation
The magnification of a concave mirror is given by:
$$m = -\dfrac{f}{f-u} $$
Now, for the given case, $$u$$ is the object distance which is equal to: $$u = f + x$$
Substituting this value in the expression for magnification, we get:
$$m = -\dfrac{f}{f - (f + x)} $$
This, gives us the value of $$m$$ to be $$ \dfrac{-f}{x} $$
Sunlight takes $$8 \ min \ 20 \ s$$ to reach the earth. If the entire space in between is filled with water, the light will take:
$$\left( _{a}\mu_{w}=\dfrac{4}{3} \right)$$
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6 min
0%
11 min 6 s
0%
8 min 10 s
0%
8 min
Explanation
Given:
Refractive index of water with respect to air $$_a\mu_w=\dfrac{4}{3}$$
$$_a\mu_w=\dfrac{\text{speed of light in air}}{\text{speed of light in water}}=\dfrac{c}{v}$$
$$\Rightarrow \dfrac{4}{3}=\dfrac{c}{v}$$
$$\Rightarrow v=\dfrac{3c}{4}$$ is the speed of light in water
With speed $$c$$, the light takes time $$t=8 \ min \ 20 \ sec$$
$$=(8 \times 60) \ sec + 20 \ sec=500 \ sec$$
Let the distance between sun and earth be $$D.$$
So the distance $$D$$, speed $$c$$ and time $$t$$ are related as
$$D= c \times t= (500) \times c$$ . . . (i)
If the speed of light becomes $$v= \dfrac{3c}{4},$$ then the time $$t'$$ taken by the light will be
$$t' = \dfrac{distance}{speed}=\dfrac{D}{\left( \dfrac{3c}{4}\right)}=\dfrac{4D}{3c}$$
Substituting value of $$D$$ from eq (i)
$$t'=\dfrac{4 (500 c)}{3c}=\dfrac{2000}{3}=666.66 \sec = 11 \ min \ 6.66 \ sec$$
$$t'\approx11 \ min \ 6 \ sec$$
So option B is the answer.
A real object is placed at a distance $$\dfrac{f}{2}$$ in front of a concave mirror having focal length $$f$$. If object is shifted by a distance $$\dfrac{f}{4}$$ away from the mirror. Find the ratio of final magnification to initial magnification.
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$$4$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{2}$$
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$$2$$
Explanation
Initially u$$=-\dfrac{f}{2}$$ ; $$ f= -f$$
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
or, $$\dfrac{1}{v}-\dfrac{2}{f}= -\dfrac{1}{f}$$
or, $$\dfrac{1}{v}=\dfrac{2}{f}-\dfrac{1}{f}$$
or, $$\dfrac{1}{v}=\dfrac{1}{f}$$
or, $$v=f$$
Initial magnification is $$-\dfrac {v}{u}=\dfrac{f}{(f/2)}$$= 2
After displacing $$u= - \dfrac{3f}{4}; f= -f $$
So, $$\dfrac{1}{v}-\dfrac{4}{3f}= -\dfrac{1}{f}$$
or, $$v=3f$$
Final magnification is $$-\dfrac{v}{u}=\dfrac{3f}{(3f/4)}=4$$
Ratio of final to initial magnification is: $$\dfrac{4}{2}= 2$$
An object is placed on the principal axis of a concave mirror at a distance of 60 cm. If the focal length of the concave mirror is 40 cm then determine the magnification of the obtained image.
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2
0%
-2
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2.5
0%
-2.5
Explanation
Focal length $${f}=-40cm$$ (-ve sign appears because the mirror is concave)
Object distance $${u}=-60cm$$
Magnification$$(m)=?$$
Using mirror formula
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{-1}{40}=\dfrac{1}{v}+\dfrac{-1}{60}$$
$$\dfrac{1}{v}=\dfrac{-3+2}{120}$$
$$v=-120cm$$
Magnification $$m=\dfrac{-v}{u}=\dfrac{120}{-60}=-2$$ (The minus sign signifies that the image is inverted)
In a vessel, as shown in fig, point P is just visible when no liquid is filled in vessel through a telescope in the air. When liquid is filled in the vessel completely, point Q is visible without moving the vessel or telescope. Find the refractive index of the liquid.
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$$\dfrac {\sqrt {14}}{3}$$
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$$\dfrac {\sqrt {85}}{5}$$
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$$\sqrt 2$$
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$$\sqrt 3$$
Explanation
When liquid is not filled in the vessel, point P can just be seen.
Thus the triangles, ABP and BCD are similar.
Thus $$\dfrac{PA}{AB}=\dfrac{BC}{CD}$$
Hence, $$CD=4R$$
Now after filling in a liquid of refractive index $$\mu$$, point Q can be seen. Hence the angle of refraction at D is $$r$$ as shown in the figure.
Applying snell's law to the refraction at point D,
$$\sin i=\mu \sin r$$
$$\dfrac{1}{\sqrt{5}}=\mu \dfrac{1}{\sqrt{17}}$$
Hence, $$\mu=\dfrac{\sqrt{85}}{5}$$
The focal length of a concave mirror is f and the distance from the object to the principal focus is p. The ratio of the size of the real image to the size of the object is:
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$$-\displaystyle \frac{f}{p}$$
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$$\displaystyle \left(\frac{f}{p}\right)^2$$
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$$\displaystyle \left(\frac{f}{p}\right)^{\frac{1}{2}}$$
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$$-\displaystyle \frac{p}{f}$$
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$$-fp$$
Explanation
Hint:
The focus of the concave mirror lies on the left side from the pole of the mirror. Thus, its focal length is negative.
Step 1: Given values.
The focal length of a concave mirror is, $$-f$$. (Concave mirror)
Distance of object from the focus of mirror is $$p$$
Thus object lies at a distance $$(p+f)$$, on the left side from the pole of the mirror.
Therefore, $$u = -(p+f)$$
Step 2: Use Mirror formula to find distance of image formed ($$v$$).
By mirror formula, we have
$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$
$$\Rightarrow \dfrac{1}{-f} = \dfrac{1}{v} + \dfrac{1}{-(p+f)}$$
$$\Rightarrow \dfrac{-1}{f} = \dfrac{1}{v} - \dfrac{1}{(p+f)}$$
$$\Rightarrow \dfrac{1}{p+f} - \dfrac{1}{f} = \dfrac{1}{v}$$
$$\Rightarrow\dfrac{1}{v} = \dfrac{f - (p+f)}{f(p+f)}$$
$$\Rightarrow\dfrac{1}{v} = \dfrac{-p}{f(p+f)}$$
$$ \Rightarrow v = \dfrac{-f(p+f)}{p}$$
Step 3 : Use formula for magnification to find required ratio.
Ih height of object is $$O$$ and the height of image formed is $$I$$, then we know that magnification is given as,
$$m = \dfrac{-v}{u} = \dfrac{I}{O}$$
$$\Rightarrow \dfrac{I}{O} = \dfrac{- \dfrac{(-f(p+f))}{p}}{-(p+f)}$$
$$\Rightarrow \dfrac{I}{O} = \dfrac{-f}{p}$$
Thus, ratio of height of image to the height of object is, $$\dfrac{-f}{p}$$.
Option A is correct.
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct
Explanation
Solution:
Let angle of incidence be $$i$$ and angle of refraction be $$r$$.
As per laws of reflection, angle of incidence is equal to angle of reflection. So, $$r_{reflection}=i $$.
Applying snell's law on the glass slab, we can write,
$$\dfrac{\sin i}{\sin r} = \dfrac{\mu}{1}$$
where $$\mu$$ is the refractive index of the glass.
Hence,
$$1\times \sin i$$ =$$\mu \sin r$$
Let reflected and refracted ray be perpendicular.
So,
$$r_{reflection}+r+90=180$$
$$i+r+90=180$$
$$\Rightarrow i+r=90$$
$$\Rightarrow r=90-i$$
So,$$\sin r= \sin (90-i)$$
So, $$\sin r= \cos i$$
Putting it in snell's law we get $$ tan\ i= \mu$$
So assertion is only true when
$$ tan\ i= \mu$$
Hence option D is correct.
When light falls on a given plate at angle of incidence of $$60^o$$, the reflected and refracted rays are found to be normal to each other. The refractive index of the material of the plate is then
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0%
0.866
0%
1.5
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1.732
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2
Explanation
Given:
Angle of incidence $$i = 60^o$$.
$$i+r = 90^o$$
$$\therefore r = 30^o$$
From Snell's law of refraction we can write,
Now $$\mu = \displaystyle \frac{sin i}{sin r} = \frac{sin 60}{sin 30} = \frac{\sqrt 3/2}{1/2} = \sqrt{3} = 1.732$$
When a light ray moves from air to a transparent medium the angle of incidence and the angle of refraction are $$45^{\circ}$$ and $$30^{\circ}$$ respectively. Find the refractive index of this medium is
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$$1.21$$
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$$1.42$$
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$$1.88$$
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$$2.42$$
Explanation
Given,
Angle of incidence $$\angle i={ 45 }^{ 0 }$$
Angle of refraction $$\angle r={ 30 }^{ 0 }$$
By Snell's law
$$n=\dfrac { sini }{ sinr } $$, where $$n$$ is the refractive index of the medium.
$$n=\dfrac { sin{ 45 }^{ 0 } }{ sin{ 30 }^{ 0 } } =\sqrt { 2 }=1.42 $$
The image of an object placed on the principal axis of a concave mirror of focal length $$12\ cm$$ is formed at a point which is $$10\ cm$$ more distant from the mirror than the object. The magnification of the image is
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$$8/3$$
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$$2.5$$
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$$2$$
0%
$$-1.5$$
Explanation
Let the object distance be $$u$$ then image distance is $$u+10$$
$$u= -u$$ ; $$v= -(u+10)$$ ; $$f= -12$$
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\dfrac{-1}{u+10}+\dfrac{-1}{u}=\dfrac{-1}{12}$$
$$\dfrac{2u+10}{u(u+10)}=\dfrac{1}{12}$$
$$u=20$$cm
$$v=-30$$cm
Magnification is $$-\dfrac{v}{u}= -\dfrac{30}{20}= -1.5$$
The refractive indices of substances P,Q, Rand S are 1.38, 1.46, 1.56 and 1.24 respectively. When light travelling in air is incident on these substances at equal angles, the angle of refraction will be maximum in
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P
0%
Q
0%
R
0%
S
Explanation
The angle of refraction($$r$$) will be the maximum when the light bends the least on changing the medium. This will happen when the refraction index of the medium is the least.
Thus, the angle of refraction will be maximum in the medium of refractive index 1.24.
A ray of light is incident from one medium $$(\mu = 1.5)$$ into another medium. If the angle of incidence and refraction are, respectively, $$60$$ and $$45$$, then what will be the refractive index of the denser medium?
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$$1.63$$
0%
$$1.43$$
0%
$$1.83$$
0%
Data insufficient
Explanation
j
Hence from Snell's law
$$\mu_{1}\sin i=\mu_{2} sin r$$
$$1.5 \sin 60^{0}=\mu_{2}\sin 45^{0}$$
$$1.5\dfrac{\sqrt{3}}{{\sqrt{2}}^{2}}=\mu_{2}\dfrac{1}{\sqrt{2}}$$
$$\mu_{2}=1.5\sqrt{\dfrac{3}{2}}$$
$$\mu_{2}=1.83$$
Which of the following correctly represents the graphical variation between very small angles of incidence (i) and refraction (r) ?
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0%
0%
0%
0%
Explanation
We know that $$\dfrac{\sin{i}}{\sin{r}}=\mu$$
For small $$i$$ and $$r$$, $$\sin{i}\approx i$$ and $$\sin{r}\approx r$$
$$\implies \dfrac{i}{r}=\mu$$
$$\implies i=\mu r$$
Hence, the graph will be a straight line passing through the origin.
Hence, the answer is option-(D).
In the figure below, PQRS denotes the path followed by a ray of light as it travels three media in succession. The absolute refractive indices of the media are $$\mu_1 \, \mu_2\, and\, \mu_3 $$ respectively. (The line segment RS' in the figure is parallel to PQ). Then :
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$$\mu_1\, > \, \mu_2\, >\, \mu_3$$
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$$\mu_1\, < \, \mu_3\, \mu_2$$
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$$\mu_1\, = \, \mu_3\, <\, \mu_2$$
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$$\mu_1\, < \, \mu_3\, <\, \mu_2$$
Explanation
From the given figure we can analyse $$\mu_2>\mu_1 ; \mu_3<\mu_2 ; \mu_1<\mu_3$$
so, $$\mu_1<\mu_3<\mu_2$$
An object is placed at a distance of 20 cm from a convex mirror or radius of curvature 40 cm. At what distance from the object should a plane mirror be placed so that the images due to the mirror and the plane mirror are in the same plane?
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15 cm
0%
30 cm
0%
60 cm
0%
40 cm
Explanation
Given: An object is placed at a distance of 20 cm from a convex mirror or radius of curvature 40 cm.
To find the distance at which the plane mirror should be placed such that the images due to the mirror and the plane mirror are in the same plane
Solution:
According to the given criteria,
Object distance from convex mirror, $$u=-20cm$$
Radius of curvature of convex mirror, $$R=40cm$$
Focal length of the convex mirror, $$f=\dfrac R2=\dfrac {40}2=20cm$$
Now applying the mirror formula we get,
$$\dfrac 1f=\dfrac 1v+\dfrac 1u\\\implies \dfrac 1{20}=\dfrac 1v+\dfrac 1{-20}\\\implies \dfrac 1v=\dfrac 1{20}+\dfrac 1{20}\\\implies v=10cm$$
So the distance is formed on the opposite side of the object at a distance of 10 cm from the mirror as shown in above fig.
The distance between object and image is $$u+v=20+10=30cm$$
When plane mirror is placed in front of object and in order to get image at same position as convex mirror, the mirror should be place at
$$\dfrac {30}2=15cm$$, as in plane mirror the object distance from mirror is equal to image distance from the plane mirror (refer above fig).
Hence the distance at which the plane mirror should be placed such that the images due to the mirror and the plane mirror are in the same plane = 15cm
What is a magnifying glass used for?
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Read large prints
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Read small prints
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To diverge rays
0%
As a diverging lens
Explanation
A magnifying glass is nothing but a convex lens used in such a way that it produces magnified image when object is viewed by placing the magnifying glass closer to object.
It is used to see and read the small letters and figures. It is used by watch makers to see the small parts and screws of the watch. Because of magnified image formed it become easy to see small things.
An object is placed on the principal axis of a concave mirror at a distance of $$60\ cm$$. If the focal length of the concave mirror is $$40\ cm$$ then determine the magnification of the obtained image.
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0%
$$4$$
0%
$$-2$$
0%
$$-4$$
0%
$$+2$$
Explanation
Given:
Distance of object $$u=-60\ cm$$
Focal length $$f=-40\ cm$$
From mirror formula
$$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$
$$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$$
$$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$$
$$v=-120 cm$$
Hence magnification is given by
$$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$$
$$m=-2$$
Which of the following is a spherical lens?
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Concave lens
0%
Convex lens
0%
Diverging lens
0%
All
Explanation
All of the above lenses have one or both surface spherical and hence are spherical lenses.
Answer-(D).
A glass slab has two long parallel faces. Light ray is incident on the glass slab and trace of light ray is drawn. Now glass slab is shifted parallel to the parallel faces. What will be the effect on the trace of light?
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It will shift laterally
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Angle of deviation will change
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No effect
0%
Depends on amount of shift
Explanation
Since thickness, angle of incidence and refractive index is same, there is no change in path of light ray.
A ray of light is incident on the interface between water and glass at an angle i and refracted parallel to the water surface, then value of $$\mu_g$$, refractive index of glass will be:
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0%
$$(4/3)\sin i$$
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$$\displaystyle\frac{1}{\sin i}$$
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$$\displaystyle\frac{4}{3}$$
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$$1$$
Explanation
Applying Snell's Law at glass-water interface,
$$\mu_g \sin i=\dfrac{4}{3}\sin r$$
Applying Snell's Law at water-air surface,
The angle of refraction at glass-water interface will be same as angle of incidence at air-water interface.
$$\dfrac{4}{3}\sin r=1\sin 90^{\circ}=1$$
$$\implies \mu_g \sin i=1$$
$$\implies \mu_g=\dfrac{1}{\sin i}$$
A spherical lens has a focal length $$2\ cm$$. The lens will be
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0%
Concave lens
0%
Convex lens
0%
Concavo convex lens
0%
None
Explanation
Concave lens has negative and concavo-convex lens can have negative focal length.
But, convex lens always has positive focal length.
Answer-(B).
The diameter of the reflecting surface of spherical mirror is called ..............
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Focus
0%
Aperture
0%
Centre of curvature
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None of these
Explanation
The diameter of the reflecting surface of the spherical mirror is called it's $$aperture$$.
Magnification of lens is:
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$$\dfrac {\text {Size of the image}}{\text {Size of the object}}$$
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$$\dfrac {\text {Image distance}}{\text {Object distance}}$$
0%
Both A and B
0%
None
Explanation
For lens, magnification is given by
$$m=\dfrac{h_i}{h_o}=\dfrac{v}{u}$$
where $$h_i$$ = image height $$h_0$$ = object height $$v$$=image distance $$u$$ = object distance
Answer-(c)
The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called ...............
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0%
Centre of curvature
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Radius of curvature
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Focus
0%
Pole
Explanation
The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror. Here, the distance $$R$$ is the radius of curvature in the figure.
Answer-(B).
Curved mirror whose reflecting surface faces towards the centre of the sphere, from which is considered to be a part of, is known as
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0%
Convex mirror
0%
Concave mirror
0%
Plane mirror
0%
None
Explanation
The given picture shows the mirror of aforesaid description and this is a concave mirror.
A water film is formed on a glass-block. A light ray is incident on water film from air at an angle of $$\displaystyle { 60 }^{ \circ }$$ with the normal. The angle of incidence on glass slab is
($$\displaystyle { \mu }_{ g } = 1.5,{ \mu }_{ w } = \frac { 4 }{ 3 } $$)
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$$\displaystyle { sin }^{ -1 }\left( \frac { 3\sqrt { 3 } }{ 8 } \right) $$
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$$\displaystyle { sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) $$
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$$\displaystyle { sin }^{ -1 }\left( \frac { 4\sqrt { 3 } }{ 9 } \right) $$
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$$\displaystyle { sin }^{ -1 }\left( \frac { 9\sqrt { 3 } }{ 16 } \right) $$
Explanation
The angle on incidence on glass slab will be equal to the angle of refraction on the air water interface.
Applying Snell's Law on air water interface, we get
$$\mu_asin \ i = \mu_wsin \ r$$
$$1 \times sin \ { 60 }^{ 0 }=\dfrac { 4 }{ 3 } sin \ r$$
$$sin \ r =\dfrac { 3\sqrt { 3 } }{ 8 } $$
$$r ={ sin }^{ -1 } \dfrac { 3\sqrt { 3 } }{ 8 } $$
As shown in above figure, a beam of light in air is incident upon the smooth surface of glass $$30^o$$ with glass surface. Calculate the refractive index of the glass if the reflected beam and refracted beam are perpendicular to each other.
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$$\frac{1}{2}$$
0%
$$\frac{1}{2}\sqrt{3}$$
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$$\sqrt{3}$$
0%
$$2\sqrt{3}$$
0%
$$3\sqrt{3}$$
Explanation
From figure, $$i=90^{o}-30^{o}=60^{o}$$
And, from law of reflection , angle of reflection=$$r=i=60^{o}$$
Since, reflected and refracted ray are normal,
hence,
$$r+R+90^{o}=180^{o}$$ where $$R$$= angle of refraction
$$\implies R=90^{o}-r=90^{o}-60^{o}=30^{o}$$
From Snell's Law:-
$$\mu=\dfrac{\sin i}{\sin R}=\dfrac{\sin 60^{o}}{\sin 30^{o}}$$
$$\implies \mu=\sqrt{3}$$
Answer-(C)
An object is placed at 20 cm in front of a concave mirror produces three times magnified real image. What is focal length of the concave mirror?
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15 cm
0%
6.6 cm
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10 cm
0%
7.5 cm
Explanation
Given : Object distance, $$u = -20$$ cm
A concave mirror forms a real magnified image only when the image to be formed is inverted i.e. $$m = -3$$
Magnification $$m = \dfrac{-v}{u}$$
$$\therefore$$ $$-3 = \dfrac{-v}{-20}$$ $$\implies v = -60$$ cm
Using mirror formula : $$\dfrac{1}{v}+\dfrac{1}{u} =\dfrac{1}{f}$$
$$\therefore$$
$$\dfrac{1}{-60}+\dfrac{1}{-20} =\dfrac{1}{f}$$
$$\implies$$ $$f = \dfrac{-60}{4} = -15$$ cm
Thus focal length of the mirror is 15 cm.
Which lens is used in a magnifying glass?
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Concave
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Convex
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Diverging
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Converging
Explanation
In a magnifying glass, a convex lens is used that is also called converging lens, thus options $$B&D$$ are correct.
A beam of light striking the surface of a glass plate from the air as shown in above figure.
If the angle between reflected beam and refracted beam is 90 degree, Find out the refractive index of the glass?
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$$\sin{{55}^{o}}$$
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$$\dfrac{1}{\sin{{55}^{o}}}$$
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$$\dfrac{1}{\sin{{35}^{o}}}$$
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$$\dfrac{\sin{{55}^{o}}}{\sin{{35}^{o}}}$$
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$$\dfrac{\sin{{35}^{o}}}{\sin{{55}^{o}}}$$
Explanation
According to figure, $$\theta_1 + 35^o =90^o$$ $$\implies \theta_1 = 55^o$$
As the angle of incidence is equal to angle of reflection, thus $$\theta_2 =\theta_1 = 55^o$$
Also $$\theta_3 = 90^o - \theta_2 = 90^o - 55^o =35^o$$
Given : $$\theta_3 + \theta_4 = 90^o$$
$$\therefore$$ $$\theta_4 =90^o - 35^o = 55^o$$
$$\implies$$ $$\theta_5 =90^o - \theta_4 = 90^o - 55^o = 35^o$$
Using Snell's law of refraction : $$n_a \times sin \theta_1 = n_g \times sin\theta_5$$ where $$n_a = 1$$
$$\therefore$$ $$1 \times sin55^o =n_g \times sin35^o$$ $$\implies n_g = \dfrac{sin55^o}{sin 35^o}$$
In above shown figure, an object is kept 20 cm from the concave mirror and formed the real image at the distance of 5 cm from the mirror. Calculate the focal length of the mirror.
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2.5 cm
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4 cm
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25 cm
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20 cm
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40 cm
Explanation
Given :
$$u = -20$$ cm
$$v = -5 $$ cm
Using Mirror equation
$$\dfrac{1}{u}+ \dfrac{1}{v} =\dfrac{1}{f}$$
$$\therefore$$
$$\dfrac{1}{-20}+ \dfrac{1}{-5} =\dfrac{1}{f}$$ $$\implies f = -4$$ cm
Thus focal length of concave mirror is $$4$$ cm.
Below is a drawing made by a student that represents a light ray entering a piece of plastic and passing into the plastic. The student drew the original light ray to enter the plastic at a $$60.0$$ degree angle with the normal as pictured.
He was able to look through the plastic and line up a ruler with the original light ray he drew to measure the angle of the light ray in the plastic. This angle is labeled in the drawing. The student used his drawing to determine the index of refraction for the plastic.
What value did he determine?
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$$1.66$$
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$$0.602$$
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$$1.47$$
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$$0.68$$
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$$1.38$$
Explanation
Here
$$i=60^{\circ}$$
$$r=36.1^{\circ}$$
From
$$n_a\sin i= n_p\sin r $$
$$1\sin 60^{\circ} = n_p\sin 36.1^{\circ} $$
$$1\times \frac{\sqrt{3}}{2}=n_p\times\sin 36.1^{\circ} $$
$$\sin 36.1^{\circ}=0.59$$
$$1\times \frac{\sqrt{3}}{2}=n_p\times 0.59 $$
$$ n_p=1.47$$
A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45$$^o$$ and that of refraction is 30$$^o$$. Find the refractive index of the medium.
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2
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$$\dfrac{1}{\sqrt 2}$$
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$$2 \sqrt 2$$
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$$\sqrt 2$$
Explanation
Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by
$$\displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r}$$
The initial medium is air. Thus $$\mu_1 = 1$$
Given incidence angle $$i = 45^o$$ and refraction angle is $$30^o$$
By Snell's law, $$\mu_2 = \mu_1\times \displaystyle \frac{\sin i}{\sin r} = 1\times \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$$
Which of the following is Snell's law
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$$n_1 sini = n_2 sinr$$
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$$n_{1}/n_{2} = \sin r \times constant$$
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$$n_{2}/n_{1} = \sin r/ \sin i$$
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$$n_{2} \sin i = constant$$
Explanation
According to Snell's Law:-
$$\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}$$ when ray of light is incident from medium-1
$$\implies n_1\sin i=n_2\sin r$$
Answer-(A)
A concave mirror of focal length 'f' produces an image 'n' times the size of the object. If the image is real then the d distance of the object from the mirror is
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$$(n + 1 )f$$
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$$\dfrac{(n+1) f}{n}$$
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$$\dfrac{(n-1) f}{n}$$
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$$(n-1)f$$
Explanation
Since the image is real, it must be inverted and magnification ratio should be negative.
Mgnification $$ = \dfrac{-v}{u}$$
$$-n=\dfrac{-v}{d}$$ , $$u=d$$ is the distance of object and and image distance is $$v$$
$$v=nd$$
Now using mirror formula
$$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$$
$$\dfrac{1}{f}=\dfrac{1}{nd}+\dfrac{1}{d}$$
So $$d=\dfrac{f(n+1)}{n}$$
A light is travelling from air into a medium. Velocity of light in a medium is reduced to $$0.75$$ times the velocity in air. Assume that angle of incidence '$$i$$' is very small, the deviation of the ray is
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$$i$$
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$$\dfrac { i }{ 3 } $$
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$$\dfrac { i }{ 4 } $$
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$$\dfrac { 3i }{ 4 } $$
Explanation
Let speed of light in air be $$c$$.
Thus speed of light in medium $$v = \dfrac{3}{4}c$$
Using $$c\sin r = v\sin i$$
As $$i$$ and $$r$$ are very very small.
$$\therefore$$ $$c \times r = \dfrac{3}{4}c\times i$$
We get $$r = \dfrac{3i}{4}$$
Deviation of the ray $$\delta = i- r$$
$$\implies$$ $$\delta = i-\dfrac{3i}{4} = \dfrac{i}{4}$$
Study the given ray diagrams and select the correct statement from the following:
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Device X is a concave mirror and device Y is a convex lens, whose focal lengths are 20cm and 25cm respectively.
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Device X is a convex lens and device Y is a concave mirror, whose focal lengths are 20cm and 25cm respectively.
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Device X is a concave lens and device Y is a convex mirror, whose focal lengths are 20cm and 25cm respectively.
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Device X is a concave lens and device Y is a concave mirror, whose focal lengths are 20cm and 25cm respectively.
Explanation
In the first case,
The ray of light is converging, when passing from lens X, so it is a convex lens.
In the second case, the ray of light is diverging, when getting reflected from the mirror Y, so it is a concave mirror.
A convex mirror of focal length $$f$$ forms an image which is $$\cfrac { 1 }{ n } $$ times the object. The distance of the object from the mirror is :
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$$\left( \cfrac { n-1 }{ n } \right) f$$
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$$\left( \cfrac { n+1 }{ n } \right) f$$
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$$(n+1)f$$
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$$(n-1)f$$
Explanation
Given,
Magnification, $$m=\cfrac { 1 }{ n } $$
So, $$\cfrac { 1 }{ n } =-\cfrac { v }{ u } $$
or $$v=-\cfrac { u }{ n } $$
By using mirror formula, $$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } +\cfrac { 1 }{ u } $$
$$=\cfrac { 1 }{ -u/n } +\cfrac { 1 }{ u } $$
$$\Rightarrow u=-(n-1)f$$
A diverging lens forms _______ & ________ image
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Erect
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Real
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Virtual
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Magnified
Explanation
The image formed by a concave lens or a diverging lens is always virtual, erect and diminished thus options $$A&C$$ is correct.
Gita arranges some mirrors in group- $$1$$ and group- $$2$$ according to the magnification, type of image and size of image from the following data.
Sr. No.
Group-$$1$$
Group-$$2$$
$$(1)$$
$$<1$$ and negative
Real, inverted and small
$$(2)$$
$$>1$$ and positive
Real, inverted and enlarged
$$(3)$$
$$>1$$ and negative
Virtual, erect and enlarged
$$(4)$$
$$<1$$ and positive
Virtual, erect and small
For which of the pairs from the group $$1$$ and $$2$$ you disagree.
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$$1$$ and $$3$$
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$$2$$ and $$3$$
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$$3$$ and $$4$$
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$$1$$ and $$2$$
Explanation
If the magnification, $$m >1$$ and positive, then the image formed is virtual, erect and enlarged in nature. Also, when $$m>1$$ and negative, then the image formed is real, inverted and magnified in nature.
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Practice Class 10 Physics Quiz Questions and Answers
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